If \(\mathbf{f} = f_1(y, z) \hat{i + f_2(z, x) \hat{j} + f_3(x, y) \hat{k}}\), then \( \mathbf{f} \) is
We are given the vector field \( \mathbf{f} = f_1(y, z) \hat{i} + f_2(z, x) \hat{j} + f_3(x, y) \hat{k} \), where the components depend on the respective variables as shown. To determine the nature of the field, we need to check if it is:
Let's calculate the divergence \( \nabla \cdot \mathbf{f} \):
\[ \nabla \cdot \mathbf{f} = \frac{\partial f_1}{\partial x} + \frac{\partial f_2}{\partial y} + \frac{\partial f_3}{\partial z} \]
Given that \( f_1 \), \( f_2 \), and \( f_3 \) depend on different variables, the partial derivatives will simplify to the following:
\[ \frac{\partial f_1}{\partial x} = 0, \quad \frac{\partial f_2}{\partial y} = 0, \quad \frac{\partial f_3}{\partial z} = 0 \]
Thus, the divergence is zero:
\[ \nabla \cdot \mathbf{f} = 0 \]
Since the divergence of \( \mathbf{f} \) is zero, \( \mathbf{f} \) is a solenoidal field.
The field \( \mathbf{f} \) is solenoidal, and thus the correct answer is option (B).
Let \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) be three vectors such that \(\mathbf{a} \times \mathbf{b} = \mathbf{a} \times \mathbf{c}\) and \(\mathbf{a} \times \mathbf{b} \neq 0. Show \;that \;\mathbf{b} = \mathbf{c}\).
If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ such that $|\vec{a}| = 3$, $|\vec{b}| = 5$, $|\vec{c}| = 7$, then find the angle between $\vec{a}$ and $\vec{b}$.
The respective values of \( |\vec{a}| \) and} \( |\vec{b}| \), if given \[ (\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = 512 \quad \text{and} \quad |\vec{a}| = 3 |\vec{b}|, \] are: